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Gh. Asachi Technical University of Iai Structural Statics Statically Determinate Structures

Department of Structural Mechanics Senior Lecturer Cezar Aanici, Dr. Eng.

- 1 -

STATICALLY DETERMINATE ARCHES

THE PARABOLIC ARCH

Let solve the following statically determinate parabolic arch:

Y

X

20kN/m

f=

5m

L/2 = 10 mL/2 = 10 m

A

C

B

H =100 kNA

V =150 kNA

H =100 kNB

V =50 kNB

L = 20 m

H =0 kNA

V =150 kNA V =50 kNB00

20kN/m

C

+

a=2ma=2m a=2m a=2m a=2m a=2m a=2m a=2m a=2ma=2m

1

2

34 5

6

7

8

-

Q0

M0

1

50

1

10

70

30

10

50

50

50

50

50

50

260

440

540

M(x)=562.5

560

500

400

300

200

100

x=7.5m

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- 2 -

1. The degree of static determinacy:

3 ( ) 3 2 (1 2 4) 0s

n c l r = + = + =

2. The equation of the parabolic arch is (see the Appendix):

2

4( ) ( )

f xy x l x

l

=

3. From the static equilibrium equations sets the reactive forces of the parabolic arch are:

( )0 : 20 20 10 5 0 50

0 : 10 5 0 100

A B B

right

C B B B

M V V kN

M V H H kN

= = =

= = =

( ) 0 : 20 20 10 15 0 150

0 : 10 20 10 5 5 0 100

B A A

left

C A B A

M V V kN

M V H H kN = = =

= = =

BecauseHA=HB, it could be generically written that the arch thrusts areHA=HB=H.

Global force equlibrium check:

0 : 100 100 0

0 : 20 10 150 50 200 0

X A B

Y A A

F H H

F V V

= = =

= + = + =

4. The recurrence relationships for the internal efforts are:

0

tan

0

0

cos sin sin cos sin

sin cos cos sin cos

i

left

i A i xi i i i i i

left

n i A i xi i i i i i

left

xi A i xi k i i i

F H V R H Q N

F H V R H Q Q

M V x R a H y M H y M

= + = =

= + = + =

= = =

where Mi0 and Qi

0are the corresponding internal efforts in the attached simple supported beam

identically loaded.

It could be observed that:

00 0 C

C C

MM M H f H

f= = =

which means that in order to reduce the thrusts magnitudes, the rise of the arch has to be decreased.

To find the internal efforts, the arch must be separated in an appropriate number of voussoirs,

the optimum number being between 8 and 12, having equal length projections on the abscissa. If the

number of the cross-sections are under the minimum recommended number, the risk of fake internal

effort diagrams is greater. Choosing more than 12 cross-sections, has frequently been observed as not

usually necessary.

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- 3 -

Y

X

f=

5m

A

H

V

x

i

i

i

n tan

HcosHsin

V sin

V cosA

A

A

i

i

i

i

yi

Rxi

ak

R cosxi i

R sinxi i

Internal efforts in

the attachedsimple

supported beam

Arch internal efforts

Cross-

section x i

[m]yi

[m]tg i sin i cos i

Qi0 Mi

0 Ni Qi Mi

A 0 - 1.000 0.707 0.707 150 0 -176.78 35.36 0

1 2 1.80 0.800 0.625 0.781 110 260 -146.80 23.43 80

2 4 3.20 0.600 0.514 0.857 70 440 -121.76 8.57 120

3 6 4.20 0.400 0.371 0.928 30 540 -103.99 -9.28 120

4 8 4.80 0.200 0.196 0.981 -10 560 -96.10 -29.42 80

C 10 5.00 0.000 0.000 1.000 -50 500 -100.00 -50.00 0

5 12 4.80 -0.200 -0.196 0.981 -50 400 -107.86 -29.42 -806 14 4.20 -0.400 -0.371 0.928 -50 300 -111.42 -9.28 -120

7 16 3.20 -0.600 -0.514 0.857 -50 200 -111.47 8.57 -120

8 18 1.80 -0.800 -0.625 0.781 -50 100 -109.32 23.43 -80

B 20 - -1.000 -0.707 0.707 -50 0 -106.07 35.36 0

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- 4 -

FINAL INTERNAL EFFORTS DIAGRAMS

A

C

B

1

2

34 5

6

7

8

-

177

147

122

10496 100 108

111

111

109

106

N [kN]

-

35

239

9 29 50 29 9

923

35Q [kN]

+ +

A

C

B

1

2

34 5

6

7

8

A

C

B

1

2

3

4

56

7

8

80

120

120

80

80120

120

80

M [kNm]

+

-

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- 5 -

Appendix

To define a parabola, three (x, y) sets of coordinates are required. For a given sag f (theheight of the arch) and

a span l(the arch width), and considering the parabolic arch with the notations above presented having the

origin of the global system of axes in the left arch spring , then three points could be identified as follows:

[ ] (0, 0)A = , [ ] , 02

lB

=

, and [ ] ,2

lC f

=

(1)

The general form of a parabola equation is:2y ax bx c= + + (2)

To find the variables a, b and c, respectively, the boundary conditions must be substituted into the above

equation of the parabola, resulting:2 0 0 0 0for x y a b c= = + + = (3a)

2

2 2 2

l l lfor x y a b c f

= = + + =

(3b)

2 0for x l y a l b l c= = + + = (3c)

The equation (3.a) obviously gives c=0, and from the equation system (3.b) and (3.c), the parabolic arch

equation is obtained as follows:

( )2

4( )

fy x x l x

l

= (4)

The slope of an i cross section of the arch is given by:

( )2( ) 4tan 2i dy x f l xdx l = = (5)

and

2

tansin

1 tan

ii

i

=+

2

1cos

1 tani

i

=+

(6)

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- 6 -

THE CIRCULAR ARCH

Let solve the following statically determinate circular arch:

Y

X

20kN/m

f=

R=

10m

R = 10 mR = 10 m

A

C

B

H =50 kNA

V =150 kNA

H =50 kNB

V =50 kNB

L = 20 m

H =0 kNA

V =150 kNA V =50 kNB00

20kN/m

C

+

1

2

3 4

5

6

-Q

0

M0

150

110

70

30

10 50 50 50 50 50 50

260

440

540

M(x)=562.5

560

500

400

300

200

100

x=7.5m

a=30

a=45

a=60

R cos a R cos a

a=30

a=45

a=60

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- 7 -


3 ( ) 3 2 (1 2 4) 0sn c l r = + = + =

2. The equation of the circular arch is:

2 2( ) ( )y x R x R=

3. From the static equilibrium equations sets the reactive forces of the parabolic arch are:

( ) 0 : 20 20 10 5 0 50

0 : 10 10 0 50

A B B

right

C B B B

M V V kN

M V H H kN

= = =

= = =

( ) 0 : 20 20 10 15 0 150

0 : 10 20 10 5 10 0 50

B A A

left

C A B A

M V V kN

M V H H kN

= = =

= = =



0 : 100 100 0

0 : 20 10 150 50 200 0

X A B

Y A A

F H H

F V V

= = =

= + = + =

4. The recurrence relationships for the internal efforts are the same as for the parabolic archas follows:

0

tan

0

0

cos sin sin cos sin

sin cos cos sin cos

i

left

i A i xi i i i i i

left

n i A i xi i i i i i

left

xi A i xi k i i i

F H V R H Q N

F H V R H Q Q

M V x R a H y M H y M

= + = =

= + = + =

= = =

where Mi0 and Qi

0are the corresponding internal efforts in the attached simple supported beam

identically loaded, the distancesxi are

- for voussoirs 1, 2, 3: xi=R(1-cos a)- for voussoirs 4, 5, 6: xi=R(1+cos a).

The slope is given by the first derivative of the arch equation:

2 2

( )tan

( )i

dy x R x

dx R x R

= =

and

tan tan(90 )i a = , sin sin(90 )i a = , cos cos(90 )i a = .

To find the internal efforts, the arch was separated into a number of 8 voussoirs, and in the following

table the internal efforts are solved using the recurrence relationships above mentioned.

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- 8 -

Internal effortsin the attached

simplesupported beam

Arch in ternal efforts

Cross-

section

Xi[m]

Yi[m]

tg i sin i cos i

Q0 M0 Ni Qi Mi

A 0.000 0 1.000 0.000 150 0 -150.00 -50.00 0.00

1 1.340 5.00 1.732 0.866 0.500 123 184.2 -131.52 18.20 -65.80

2 2.929 7.07 1.000 0.707 0.707 91.4 353.65 -99.98 29.27 0.10

3 5.000 8.66 0.577 0.500 0.866 50 500 -68.30 18.30 66.99

C 10.000 10.00 0.000 0.000 1.000 -50 500 -50.00 -50.00 0.00

4 15.000 8.66 -0.577 -0.500 0.866 -50 250 -68.30 -18.30 -183.01

5 17.071 7.07 -1.000 -0.707 0.707 -50 146.5 -70.71 0.00 -207.05

6 18.660 5.00 -1.732 -0.866 0.500 -50 67.5 -68.30 18.30 -182.50

B 20.000 0 -1.000 0.000 -50 0 -50.00 50.00 0.00

FINAL INTERNAL EFFORTS DIAGRAMS

A

C

B

1

2

3 4

5

6

-

150

131.5

99.98

68.3

50

68.3

70.7

50

N [kN]

68.3

-A

C

B

1

2

3

4

5

6

-

50

18.20

29.27

18.350

50

Q [kN]-

A

C

B

1

2

3

4

5

6

0.1

65.8

67

183.1

207

M [kNm]

182.5

+

+

+

18.3

18.3

0

+

-

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- 9 -

THE ARCH WITH OVERHEIGHTED TIE ROD

Let find the internal efforts in the tie rods for the following statically determinate parabolic arch:

Y

X

20kN/m

f=

5m

L/2 = 10 mL/2 = 10 m

A

C

BH =0A

V =150 kNA V =50 kNB

1

2

34 5

6

7

8T T

J J'

a

a


3 ( ) 3 7 (9 2 3) 0s

n c l r = + = + = 2. The equation of the parabolic arch is (see the Appendix):

2

4( ) ( )

f xy x l x

l

=

3. From the static equilibrium equations sets the reactive forces of the parabolic arch are (it isa simply supported undeformable structure):

0 : 0X A

F H= = ( ) 0 : 20 20 10 5 0 50A B BM V V kN= = = ( )

0 : 20 20 10 15 0 150B A A

M V V kN= = =



0 : 20 10 150 50 200 0Y A A

F V V= + = + = 4. Making a complete cross-section (a-a) through the internal hinge C of the arch, and

replacing the tie rod with a couple of internal efforts T, the following equations could be

written as follows:

0 : 10 3 0 166.67rightC BM V T T kN= = =

or

0 : 10 20 10 5 3 0 166.67leftC AM V T T kN= = =

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- 10 -

5. The efforts in the inclined and vertical tie rods, respectively, are determined usingthe joint isolation method:

T=166.67 kN

J

NJ3

NJA

18

0 : cos18 0 158.51X JA JAF T N N kN = = = 3 3

0 : sin18 0 51.50Y J J

F T N N kN = = =

Consquently, the arch could be considered as being subjected to the following equivalent loadsystem of forces:

Y

X

20kN/m

f=

5m

L/2 = 10 mL/2 = 10 m

A

C

B

V =150 kNA V =50 kNB

1

2

34 5

6

7

851.50 kN

158.51kN

51.50 kN

158.51kN

or, after decomposing on both directions the inclined forces and adding them with the

vertical reactive forces,

Y

X

20kN/m

f=

5m

L/2 = 10 mL/2 = 10 m

A

C

B

V =198.98 kNA V =98.98 kNB

1

2

34 5

6

7

851.50 kN

150.75 kN

51.50 kN

150.75 kN

and the same recurrence relationships could be used to solve for the internal efforts diagrams,

but being careful that the concentrated forces will determine jumps in the axial force and

shear force diagrams, respectively.

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